Intreat me not
to leave thee, or to return from following after thee, for whither thou
goest, I will go, and where thou lodgest, I will lodge. Thy people shall be
my people, and thy God my God. Where thou diest, will I die, and there will I
be buried. The Lord do so to me, and more also, if ought but death part thee
and me.

Ruth,
1:16

A
round-robin tournament involving N teams consists of N(N−1)/2 matches
such that each team plays each other team exactly once. Let T(N) denote the
number of distinct sets of outcomes of a round-robin tournament of N teams in
which each team wins exactly half of its games. Obviously if N is even then
each team plays an odd number of other teams, so no team can win exactly half
its games. Thus we can assume N is odd (although we can ask more general
questions that apply to both odd and even N).

One
simple idea for characterizing the number of distinct round-robin ties for N
= 2n+1 teams is that it equals the coefficient of (x1 x2
... xN)n in the expansion of

Of
course this expansion consists of 2N(N-1)/2 terms, so it's not
feasible to write out the entire expansion just to pick off the coefficient
of one particular term. However, notice that we can explicitly define T(N) in
terms of a certain partial derivative:

For
example, with N=3 we have n=1 and

Only
the final term, 2x1x2x3, contains all three
of the variables, so if we take the partial derivative of f with respect to x1,
then x2, and then x3, we have T(3) = 2. The benefit of
this approach is that we can evaluate the partials of f without actually
expanding the product, and we can perform this evaluation in a way that takes
advantage of symmetry, so that the number of non-zero terms is relatively
small.

To
see how this approach can be developed into useful formulas, first consider
again the trivial case N=3. I'll use x,y,z in place of x1, x2,
x3 for typographical convenience. We have the fundamental
function

and
we want to evaluate the partial with respect to x (at x=0), which can be
represented by

in
the limit as q goes to zero. Now we want the partial of this function with
respect to y (at y=0), which we can represent by the limit of

as
q goes to zero. Lastly, we want the partial of this function with respect to
z (at z=0), which is

If
we substitute all the way back to give an expression for fxyz in
terms of the fundamental function f, we have

The
coefficient of the general term f(r,s,t) is the product M(r)M(s)M(t) where
M(q) = 1 and M(−q) = −1. (We'll see how this generalizes below.) Also,
recalling that f(x,y,z) = (x+y)(x+z)(y+z), it's clear that if any two of the
arguments of f sum to zero, then f must equal zero. Therefore we are left
with simply

Therefore,
the coefficient of xyz in the expansion of f(x,y,z) is 2/(1!)3 =
2.

Now
for a slightly less trivial example, consider the case N=5. In this case the fundamental
function is the product of all sums of two of the five variables v,w,x,y,z,
and T(N) is the coefficient of (vwxyz)2 in the expansion of f.
This means we need to evaluate the second partial derivatives with
respect to each of these variables. So, in order to get the second
derivatives we need to evaluate the function at three values, q, 0, and −q.
It turns out that the overall partial is a linear combination of the f
functions with every possible set of five arguments from the set {q,0,−q},
and the coefficient of the general term f(r,s,t,h,g) is the product

M(r)M(s)M(t)M(h)M(g)

where
M(q)=1, M(0) = −2, and M(−q) = 1. (As will become apparent, the
"M" functions are always just the nth row of binomial coefficients
with alternating signs.) It would be impractical to write out the entire
expression, because there are 53 = 125 terms. However, we again
notice that if any two of the arguments of f sum to zero, then f itself is
zero, so it's clear that we need not consider any terms with more than one
"0" argument, nor any that contain both q and −q. Thus, we
need only consider the terms that contain all +q arguments (possibly with
exactly one 0), or all −q arguments (possibly with exactly one 0). Thus
we have

Notice
that the "denominator" for these derivatives was just q (instead of
2q as in the case N=3), so the overall denominator of second partials with
respect to all five variables is (q2)5. Also, notice
that we've represented the f terms containing four q's and one 0 by just a
single term with a multiplier of 5, because there are 5 possible slots for
the zero to be placed, and similarly for the terms with four −q's and
one 0. So, the overall (second) partial with respect to the five variables is
simply

This
means the coefficients of (vwxyz)2 in the expansion of
f(v,w,x,y,z) is 768/(2!)5 = 24, so we have T(5) = 24.

Now
let's go on to the case N=7. Here we need to take the third partials with
respect to each of seven variables, and we do this by evaluating the
fundamental function f at values of {+3q, +q, −q, −3q}. Again we
arrive at a linear combination of f terms with every possible set of
arguments from the set {3q, q,−q,−3q}, and the coefficients are
given as products of the M values M(3q) = 1, M(q) = −3, M(-q) = 3, and
M(−3q) = −1. We can neglect any terms whose arguments include
both 3q and −3q, or both q and −q. Thus we need only consider the
terms whose arguments consist entirely of {3q,q} or {3q,−q} or {−3q,q}
or {−3q,−q}. Let's consider the general case of the terms whose
arguments consist of {r,s} where r+s is not zero. This means we could have
all seven of the arguments equal to r, or six r's and one s, or five r's and
two s's, and so on. Thus the sum of all such terms is

So
we can evaluate this sum for {r,s} equal to each of the pairs {3q,q}, {3q, −q},
{−3q,q} and {−3q,-q} to give the complete "numerator"
of the (third) partial derivative with respect to each of the seven
parameters. However, this will result in "double-bookkeeping" of
the terms that consist of just a single argument, such as {3q}. Basically,
the terms of the summation with i=0 and i=7 each occur in two of the
summations, so we have to deduct one copy of each of those terms. Then,
noting that the "denominator" of the overall third partial of seven
variables with increment 2q is ((2q)3)7, we arrive at

and
so T(7) = 2640.

To
conclude with a non-trivial example, let's consider the case N=9. In this
case we need to take the 4th partial of f with respect to nine variables, and
we do this by evaluating f with various combinations of the arguments {2q, q,
0,−q,−2q}, for which the "coefficient factors" are

Again
we see that the only non-zero values of f will occur for terms whose
arguments are from the set {2q,q,[0]}, {2q,−q,[0]}, {−2q,q,[0]},
or {−2q,−q,[0]}, where [0] signifies that there can be at most
one argument equal to 0. By the same reasoning as before, the sum of the f
terms containing no "0" argument is given by

and
the sum of the f terms containing exactly one 0 is given by

Note
that we've multiplied the second summation by 9 because there are nine
possible slots for the single 0 to be placed, and we've multiplied it by 6 to
account for the single 0 argument with M(0) = 6. Also, note that if we
evaluate these two summations for each of the sets {r,s} = {2q,q} etc., we
will double bookkeep the leading and trailing terms of each summation, so we
need to deduct one copy of each of these terms. Having done that, and noting
the "denominator” of the overall 4th partial with respect to nine
variables with an increment of q is (q4)9, we arrive at

so
we have T(9) = 3230080.

Another
way of computing T(N) is based on "n-fold derangements" of the
permutations of N variables, but we leave that for another time.